> Hey, I'd agree with this-- and it's worth noting that 17^2 - 5^2 > 16^2, so even 1MPH slower would likely have resulted in no contact in this scenario.

Only with instant reaction time and linear deceleration.

Neither of those are the case. It takes time for even a Waymo to recognize a dangerous situation and apply the brake and deceleration of vehicles is not actually linear.

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mlyle 7 hours ago | parent | next [-]

> It takes time for even a Waymo to recognize a dangerous situation

Reaction time makes the math even better here. You travel v1 * reaction_time no matter what, before entering the deceleration regime. So if v1 gets smaller, you get to spend a greater proportion of time in the deceleration regime.

> linear deceleration.

After reaction time, stopping distance is pretty close to n^2. There's weird effects at high speed (contribution from drag) and at very low speed, but they have pretty modest contributions.

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Aloisius 6 hours ago | parent [-]

I was thinking more that how hard the brakes are applied likely varies based on uncertainty of a collision.

Without that these vehicles could only start braking when certainty crossed some arbitrary threshold.

	▲	mlyle an hour ago \| parent [-]
		I think the strategy is a lot more nuanced than that. In any case, with zero reaction time, linear deceleration time to stop is proportional to velocity squared. With reaction time, the linear deceleration time is that plus the velocity times the reaction time. so the two cases we're comparing are 17 * r + (17^2 - 5^2) vs. 16 * r + (16^2), or 17 * r + 264 vs 16 * r + 256. As long as reaction time isn't negative, a vehicle that could slow to 5MPH starting at 17MPH could slow to 0MPH starting at 16MPH. (There are weird things that happen at <2.5MPH reducing deceleration, but the car moves only a few inches at these speeds during a panic stop).

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6 hours ago | parent | prev [-]

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